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Consider, for example, the linear congruence:

$56x \equiv 23 ($mod $93)$ if we know that the inverse of of $56$ modulo $93$ is $5$.

Multiplying both sides by the inverse, $5$, we have: $280 x \equiv 115 ($mod $93)$.

Thus: $x \equiv 115($mod $93)$.

From here on, I do not know how to solve it.

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    $\begingroup$ Reduce 115 mod 93 (115-93=?). $\endgroup$ – YoTengoUnLCD Jul 9 '15 at 21:42
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Well, you already solved it. The solution is $x\equiv115\bmod 93$. Although it seems nicer to change $115$ for the equivalent $22$ to get $x\equiv 22\bmod 93$

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  • $\begingroup$ This means that $x = 22 + k93$ for $k$ an integer, right? $\endgroup$ – Julia Jul 9 '15 at 21:46
  • $\begingroup$ Yes, the integers $x$ such that $56x$ leaves remainder $23$ when dividing by $93$ are those of the form $22+93k$ $\endgroup$ – Jorge Jul 9 '15 at 21:50

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